# The uniform topology on $\mathbb{R}^J$ is finer than the product topology and coarser than the box topology

In Munkres Topology (page 124, Ch: Metric Topology) he states the following theorem:

The uniform topology on $\mathbb{R}^J$ is finer than the product topology and coarser than the box topology; these three topologies are all different if $J$ is infinite.

By the statement of the theorem, it seems Munkres is referring to the finer and coarser comparisons as those of the finite product/box topology. However, earlier in the text he states that for a finite product space the box and product topology are the same!

So should not the uniform topology either be finer than both the product topology and box topology or coarser for both the product/box topology?

Thanks!

Note:

Uniform metric definition for an $n$ coordinate product space:

$$\bar{p}(x,y) = \sup\left\{\bar{d}(x_i - y_i), i \in n \right\}.$$

• "Finer" and "coarser" are not strict conditions. If $J$ is finite, then the product topology is both finer and coarser than the box topology (implying that they are the same). The uniform topology fits between the two, though the distinction only matters when $J$ is infinite. – Xander Henderson Dec 4 '17 at 5:35
• ah shoot! i now remember thinking this when first reading over the theorem! so to be clear, Munkres could have written the theorem switching finer and coarser and it would still hold true, correct? (in the finite case) – H_1317 Dec 4 '17 at 5:41
• In the finite case, they are all the same, hence I suppose that one could reverse the language, but that would be rather silly, as the theorem is only interesting in the infinite case, where the relation is as stated. – Xander Henderson Dec 4 '17 at 5:43
• @XanderHenderson However, the author explicitly states that if $J$ is infinite, these three topologies are different! – onurcanbektas Jan 17 '18 at 10:07
• @onurcanbektas Yes, and? Isn't that exactly what I said? – Xander Henderson Jan 17 '18 at 14:24

Let $\mathcal T_p$, $\mathcal T_b$, and $\mathcal T_u$ be the product, box, and uniform topologies on $\mathbb R^J$, respectively.
For any $J$, it is the case that $\mathcal T_p\subseteq\mathcal T_u\subseteq\mathcal T_b$.
However, if $J$ is finite, then $\mathcal T_p=\mathcal T_b$, which implies that $\mathcal T_p=\mathcal T_u=\mathcal T_b$.
• However, the author explicitly states that if $J$ is infinite, these three topologies are different! – onurcanbektas Jan 17 '18 at 10:05